Probability donkey alive

Question

I have There was a cowboy and 3 donkeys (Donkey A, Donkey B, Donkey C). The cowboy wears an eye cover and shoots randomly. What is the probability of donkey A to still be there after the cowboy shot 2 bullets? There is equal chances that the cowboy hits Donkey A, Donkey B, Donkey C or a Miss.

I am helpless regarding this. I don't know how to solve it. My teacher asked me to solve it by finding the probability that donkey A still alive. But I want to solve it straight forward and directly. Is it possible? If yes, how?

Answer 1

Here is a tree diagram representing the situation. For example, $$\text{Pr}(\text{Hit A on First shot)}=1/4$$ To solve the problem, add up all the cases where $\text{Donkey A}$ survives. $$\text{Pr}(\text{Miss A on First shot and Second shot})$$ $$=\text{Pr}(\text{Hit B or C on First shot and Miss A on Second Shot})$$ $$+\text{Pr}(\text{Miss on First shot and Miss A on Second Shot})$$ $$=\frac{1}{2}\cdot\frac{2}{3}+\frac{1}{4}\cdot\frac{3}{4}=\frac{25}{48}\approx0.52.$$

We must split into cases because the probability of hitting $\text{Donkey A}$ on the second shot depends on the outcome of the first shot. If the first shot is a miss, there is a $1/4$ probability $\text{Donkey A}$ is shot second. However if $\text{Donkey B or C}$ was shot first, then there is a $1/3$ probability $\text{Donkey A}$ is shot second

Answer 2

As remarked by hardmath the protocol is somewhat ambiguously defined. Since the cowboy is blind he may shoot at a dead donkey the second time. I add a dummy donkey D to account for the misses. Then donkey A survives the first as well as the second shot with probability ${3\over4}$ each. Assuming these shots to be independent the probability that A is still alive at the end is ${9\over16}$.